# If a sequence diverges to infinity then so do all its subsequences. Then why does the sum for $\frac{1}{n}$ diverge but not $\frac{1}{n^2}$?

If a sequence diverges to infinity then its subsequences diverge to infinity as well.

How does the fact that the sequence of partial sums $S_n = \sum_{k=1}^{n}\frac{1}{k} \to \infty$ as $n \to \infty$, and the subsequence $S_{n^p} = \sum_{k=1}^{n}\frac{1}{k^p}$ converges when $p>1$ not violate the first proposition? Is my initial assumption wrong? Am I wrong in calling $S_{n^2}$ a subsequence?

Thank you for the help.

• Your $S_{n^p}$ is not a subsequence of $S_n$. For example, $\sum_{k=1}^2 1/k^2 = 1 + 1/4 = 5/4$, whereas the first few terms of the sequence $S_n$ are $1, 3/2, 11/6$. – Bungo Dec 1 '17 at 21:36
• $\sum_{k=1}^{n^p}\dfrac1k$ is way bigger than $\sum_{k=1}^n\dfrac1{k^p}$. – Jyrki Lahtonen Dec 1 '17 at 21:37
• You are mixing up sequences and series. – Angina Seng Dec 1 '17 at 21:37
• Put another way, you are mixing up a sequence with the corresponding sequence of its partial sums. 1/n^2 is a sub-sequence of 1/n, but the partial sums of 1/n^2 do not comprise a sub-sequence of the partial sums of 1/n. – Sridhar Ramesh Dec 1 '17 at 21:43

Yes, you are wrong. The series $\sum_{n=1}^\infty\frac1n$ diverges to $+\infty$. What this means is that the sequence$$\tag{1}1,1+\frac12,1+\frac12+\frac13,\ldots$$diverges to $+\infty$. Now consider the series $\sum_{n=1}^\infty\frac1{n^2}$. The first partial sums are$$1,1+\frac14,1+\frac14+\frac19,\ldots$$The terms of this sequence (except for the first one) don't appear in $(1)$. Therefore, it is not a subsequence of $(1)$. So, there is no contradiction.

Note that $S_{n^{p}}=\displaystyle\sum_{k=1}^{n^{p}}\dfrac{1}{k}$, so the $p$-series is not a subsequence of $\{S_{n}\}$.

A little more:

$a_n =1/n$; a subsequence of $(a_n)_{n \in \mathbb{N}}$ is $b_n =1/n^2$, $(b_n)_{n \in \mathbb{N}}$.

The sequence $a_n$ converges to $0$, so does the subsequence $b_n.$

$S_n = \sum_{k=1}^{n} a_n$ diverges.

$T_m = \sum_{k=1}^{m} b_n$ converges,

But:

$(T_m)_{m \in \mathbb{N}}$ is not a subsequence of $(S_n)_{n \in \mathbb{N}}.$